So, applying L to both sides yields

(s Y(s) 0) Y(s) = 2s/(s^2 1)^2

==> Y(s) = 2s/[(s 1) (s^2 1)^2]

Since L{sin t} = 1/(s^2 1), we have

L{t sin t} = -(d/ds) 1/(s^2 1) = 2s/(s^2 1)^2.

So, applying L to both sides yields

(s Y(s) 0) Y(s) = 2s/(s^2 1)^2

==> Y(s) = 2s/[(s 1) (s^2 1)^2]

By partial fractions,

Y(s) = (1/2) [s/(s^2 1) 1/(s^2 1) 2s/(s^2 1)^2 2/(s^2 1)^2 1/(s 1)].

Inverting yields

y = (1/2) [cos t sin t t sin t (sin t t cos t) e^(-t)]

..= (1/2) [cos t t sin t t cos t e^(-t)].

So, applying L to both sides yields

(s Y(s) 0) Y(s) = 2s/(s^2 1)^2

==> Y(s) = 2s/[(s 1) (s^2 1)^2]

This post is last updated on hrtanswers.com at Date : 1st of September – 2022